# sampling distribution of sample means

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• Dec 28th 2007, 03:00 AM
0123
sampling distribution of sample means
I beg your help on this excercice. Thank you.

The number of hours spent studying by students the week before exams is normal distributed with standard deviation 8.4 hours. A random samole is taken to estimate the population mean number of hours studying.
A) How large a sample is needed to ensure that the probability tha the sample mean differs from the population mean by more than 2 hours is less than 0.05?

B) without doing calculations, state whether a larger or smaller sample than in part A) would be required to guarantee that the probability of the sample mean differs from the population mean by more than 2 hours is less than 0.10.

C) without doing calculations, state whether a larger or smaller sample than in part A) would be required to guarantee that the probability of the sample mean differs from the population mean by more than 1.5 hours is less than 0.05.

Thank you.
• Dec 28th 2007, 09:03 AM
TKHunny
Generally, and you should know this, $n\;=\;\frac{z^{2}s^{2}}{d^{2}}$. You should be able to see this from your previous experience building confidence intervals.

After that useful forumla, B) and C) are eyeball problems. If you increase the numerator or decrease the denominator, what does that do to the required sample size? Similarly, if you decrease the numerator or increase the denominator, what does that do to the required sample size?

Note: It should be obvious that a small sample would lead to 't' rather than 'z'.

Note: With a Finite Population Correction Factor, this will be different. You tell me what that is. We don't need it for this problem. Why?
• Dec 28th 2007, 10:20 AM
0123
Quote:

Originally Posted by TKHunny
Note: With a Finite Population Correction Factor, this will be different. You tell me what that is. We don't need it for this problem. Why?

We do not know N, so how can we compute the finite population correction factor?:(

I do not understand the point A.

What I do is

P(X>p+2) = P(Z> (p+2 - p)/ (8.4/squarerootof n) )= 0.05

so I find Z=1.65 and then 1.65= 2/ (8.4/ sqr n )
n= (1.65* 8.4/2)^2 =48 but the book says 68.

I do not know. :(
• Dec 31st 2007, 10:15 AM
TKHunny
Quote:

Originally Posted by 0123
We do not know N, so how can we compute the finite population correction factor?:(

Why do we care? My point is only that if we need the FPCF, we'll have to calculate a different value for the sample size.

Quote:

I do not understand the point A.

What I do is

P(X>p+2) = P(Z> (p+2 - p)/ (8.4/squarerootof n) )= 0.05

so I find Z=1.65 and then 1.65= 2/ (8.4/ sqr n )
n= (1.65* 8.4/2)^2 =48 but the book says 68.

I do not know. :(
When you decide that it is a 2-tail test, you will have it.
• Dec 31st 2007, 10:26 AM
0123
where does it come out that it is a two tail test?
• Dec 31st 2007, 10:44 AM
TKHunny
You must read the problem very carefully.

"the probability that the sample mean differs from the population mean by more than 2 hours"

Which direction are we looking? Higher? Lower? Both? Either?
• Dec 31st 2007, 10:47 AM
0123
Quote:

Originally Posted by TKHunny
You must read the problem very carefully.

"the probability that the sample mean differs from the population mean by more than 2 hours"

Which direction are we looking? Higher? Lower? Both? Either?

(Doh)(Crying) Right. Thank you.